Let $F$ be a field of characteristic that is not $2$.
I want to prove that
$x\in\overline{F}$ is in $F(\sqrt{F})$ $\iff$ $F\subset F(x)$ is a finite Galois extension for which the group Gal$(F(x)/F)$ is abelian of exponent $2$.
What I know:
Sadly, not much. The hint I got is $F(x)\subset F(x_1,...,x_n)$ for $x_i^2\in F$.
But I don't see how I can use this.
By definition $F(\sqrt F$) is the compositum of all the quadratic extensions of $F$, or equivalently, is obtained by adding to $F$ the square roots of all the elements of $F^*$ (taken modulo $F^{*2} $ obviously). Galois theory tells you that G := Gal($F(\sqrt F) /F$) is a product of copies of $\mathbf Z/2\mathbf Z$ , an abelian group of exponent 2 . Your subextension $F(x)/F$ will have a Galois group H isomorphic to a finite quotient of G, i.e. a product of a finite number of $\mathbf Z/2\mathbf Z$ . Apply again Galois to get what you want.
To avoid calculations, you could also apply Kummer’s theory, see e.g. Lang’s « Algebra », VIII, §8. which tells you precisely that G $\cong$ Hom($F^{*}/ F^{*2}$, {$\pm1$}), hence H will be isomorphic to the Hom of a finite subgroup of $F^{*}/ F^{*2}$, and you are done.